Permutation Calculator
Free Permutation Calculator for statistics. Enter values to get step-by-step solutions with formulas and graphs. Get results you can export or share.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
P(n,r) = n! / (n-r)!
Permutations count the number of ordered sequences of length r that can be formed from n distinct items, where each item can be used at most once. It equals n × (n-1) × (n-2) × ... × (n-r+1), which simplifies to n!/(n-r)!.
Worked Examples
Example 1: Awarding medals
Problem:In how many ways can gold, silver, and bronze medals be awarded among 8 finalists?
Solution:P(8,3) = 8! / (8-3)! = 8! / 5! = 8×7×6 = 336.
Result:P(8,3) = 336 possible medal orderings
Example 2: Password without repeated characters
Problem:How many 4-character passwords can be formed from a 10-character alphabet if no character may repeat?
Solution:P(10,4) = 10! / (10-4)! = 10×9×8×7 = 5,040.
Result:P(10,4) = 5,040 possible passwords
Frequently Asked Questions
What is a permutation in mathematics?
A permutation is an ordered arrangement of r items selected from a larger set of n items, where the sequence matters. P(n,r), read 'n permute r', counts how many distinct ordered sequences of length r can be built — assigning gold, silver, and bronze medals to 3 of 8 runners is a permutation problem because swapping which runner gets which medal creates a different outcome.
What is the difference between a permutation and a combination?
A permutation counts arrangements where order matters (1st, 2nd, 3rd place finishers), while a combination counts selections where order doesn't matter (choosing any 3 team captains with no ranking). Because every group of r items can be arranged in r! different orders, P(n,r) = C(n,r) × r! — permutations are always at least as large as the matching combination count.
How is the permutation formula derived?
To fill r ordered positions from n available items: there are n choices for the first position, n-1 remaining choices for the second (since one item is used), n-2 for the third, and so on down to n-r+1 choices for the last position. Multiplying these together, n × (n-1) × (n-2) × ... × (n-r+1), simplifies to n! / (n-r)! — the standard permutation formula.
How large can permutation results get?
Even modest inputs produce enormous numbers — P(20,10), the number of ways to award 10 distinct ordered prizes among 20 contestants, is 670,442,572,800 (over 670 billion). This rapid growth is exactly why permutations are used to argue that brute-force password guessing or exhaustive search of large arrangement spaces is computationally infeasible.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy